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EXAMPLE:
Correction of the actuation time:
A motor operating at 80% In is switched off and switched
on immediately.
The starting duty is 3xIn @ 25s. The selected Thermal
Class is Class 20 with 39s @ 3xIn.
The correction factor for 80% In in the table 6.4 is 0.35.
The corrected actuation time is: 0.35 x 39s = 13.7s, the
actuation time is reduced from 39s at cold condition to
13.7s. This means that a immediate start is not possible
before the thermal image of the motor has been cooled
down.
NOTE!
For programming correctly the Thermal Class that should
protect the motor against overheating, you must consider
the allowed hot locked rotor time. This data is available in
the catalog of the motor manufacturer.
EXAMPLE:
Suggestions on how to program the Thermal Class
correctly:
Motor data:
Power: 100hp
Voltage: 380V
Rated current (In): 140A
Service factor (S. F.): 1.00
Ip/In: 6.6
Locked rotor time: 12s at hot (Standard catalog information)
Speed: 1780rpm
Data about the motor + load Starting:
Starting by Voltage Ramp, average starting current:
3 x the rated motor current during 25s (3 x In @ 25s).
1) In the cold chart Figure 6.10, we can find the
minimum required Thermal Class that allows motor start
with reduced voltage:
For 3 x In @ 25s, we select the next higher class: 15
2) In the hot chart Figure 6.11, we can find the maximum
Thermal Class that the motor will withstand due to the
hot locked rotor time:
For 6.6 x In @ 12s, we select the next lower Class: 30
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DETAILED PARAMETER DESCRIPTION
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